Question

3. If x electrons are needed to displace 108 g silver from a solution which contains Ag ions,

Answer 1

Answer:

Choice A. $x$ electrons would be required for displacing $9\; \rm g$ of aluminum from a solution of $\rm Al^{3+}$ ions.

Assumption: by "$\rm Ag$ ions" the question meant $\rm Ag^{+}$ with a charge of $+1$ on each ion.

Explanation:

The question states that the relative atomic mass of $\rm Ag$ is $108$. In other words, each mole of

Therefore, that $108\; \rm g\!$ of silver that were formed would contain $1\; \rm mol$ of silver atoms.

Metallic silver would precipitate out of this $\rm Ag^{+}$ solution only after these ions are turned into $\rm Ag$ atoms.

One $\rm Ag^{+}$ ion carries one unit of positive electrical charge. On the other hand, each  $e^{-}$ carries one unit of negative electrical charge.

Therefore, each $\rm Ag^{+}\!$ ion will need to gain one electron to form a neutral $\rm Ag$ atom.

${\rm Ag^{+}}\; (aq) + e^{-} \to {\rm Ag}\; (s)$.

At least  $1\; \rm mol$ of electrons would be required to turn $1\; \rm mol\!$ of $\rm Ag^{+}$ ions into that $1\; \rm mol\!\!$ of silver atoms (which have a mass of $108\; \rm g\!$.)

Hence, $x = 1\; \rm mol$.

Unlike $\rm Ag^{+}$ ions, each aluminum ion $\rm Al^{3+}$ carries three units of positive electrical charge. That is three times the amount of charge on one $\rm Ag^{+}\!$ ion. Therefore, three electrons will be required to turn one $\rm Al^{3+}\!$ ion to an $\rm Al$ atom.

${\rm Al^{3+}}\; (aq) + 3\, e^{-} \to {\rm Al}\; (s)$

The question states that the relative atomic mass of $\rm Al$ is $27$. Therefore, each mole of $\rm Al\!$ atoms would have a mass $27\; \rm g$. There would be $\displaystyle \frac{9\; \rm g}{27\; \rm g \cdot mol^{-1}} = \frac{1}{3} \; \rm mol$ of atoms in that $9\; \rm g$ of $\rm Al\!\!$.

It takes $3\; \rm mol$ of electrons to turn one mole of $\rm Al^{3+}$ ions to one mole of $\rm Al$ atoms. Hence, $\displaystyle \frac{1}{3}\times 3\; \rm mol = 1\; \rm mol$ of electrons would be required to produce that $\displaystyle \frac{1}{3}\; \rm mol$ of $\rm Al\!$ atoms (which has a mass of $9\; \rm g$) from $\rm Al^{3+}\!$ ions.

That corresponds to the first choice, $x$ electrons.